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(D) Let the displacement of the two particles be $x_1 = A \sin(\omega t + \phi_1)$ and $x_2 = A \sin(\omega t + \phi_2)$.At the point where they pass

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(D) Let the displacement of the two particles be $x_1 = A \sin(\omega t + \phi_1)$ and $x_2 = A \sin(\omega t + \phi_2)$.At the point where they pass each other,$x_1 = x_2 = \frac{A}{\sqrt{2}}$.For particle $1$,moving away from the mean position,$\sin(\omega t + \phi_1) = \frac{1}{\sqrt{2}}$,which gives the phase $	heta_1 = \frac{\pi}{4}$.For particle $2$,moving towards the mean position at the same displacement,$\sin(\omega t + \phi_2) = \frac{1}{\sqrt{2}}$. Since it is moving in the opposite direction,its phase must be in the second quadrant,so $	heta_2 = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.The phase difference is $\Delta \phi = |	heta_2 - 	heta_1| = |\frac{3\pi}{4} - \frac{\pi}{4}| = \frac{2\pi}{4} = \frac{\pi}{2}$.Converting to degrees,$\frac{\pi}{2} = 90^{\circ}$.

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